(navigation image)
Home American Libraries | Canadian Libraries | Universal Library | Community Texts | Project Gutenberg | Children's Library | Biodiversity Heritage Library | Additional Collections
Search: Advanced Search
Anonymous User (login or join us)
Upload
See other formats

Full text of "An Introduction to Heat Transport by Conduction and Fourier's Law"

An Introduction to Heat Transport by Conduction 
and Fourier's Law 

By Patrick Bruskiewich 

Fourier's Law 

The relationship between the heat flow and the temperature difference was discovered 
experimentally in 1804 by the scientist French scientist Biot, who discovered 
experimentally that the driving force for heat was the temperature difference AT across 
an object 



q x cc -AT 



The direction of the heat flow is from the higher temperature to the lower temperature 
(refer to Fig. 1: The Direction of Heat Flow). 



J 




r a < r £ 



Fig. 1: The Direction of Heat Flow 



The full theory of heat transport by conduction was explained by the French scientist Jean 
Baptiste Joseph Fourier (1768-1830) and first presented to the world in 1812. In the 
1812 French Science Prize competition the mathematicians Laplace, Lagrange and 
Legendre assigned the problem to French scientists to discover "The Mathematical 
Theory of the laws of the propagation of heat and the comparison of the results of this 
theory with exact experiment." 

Fourier would win the 1812 French Science Prize by deriving the expression which now 
carries his name. 

The full theory was published in book form in 1822 by Fourier in the book Theorie 
Analytic de la Chaleur which is considered on par with Newton's book Principia 
Mathematica. Theorie Analytic de la Chaleur is considered one of the Great Books of the 
Western World and even today, nearly two centuries after its publication, a book worth 
reading. 

Heat Transport by conduction is described by Fourier's Law, which in one dimension is 
given by the expression 

AU 7 A AT 

q x = — = - k A — 

At Ax 



where q x is the rate of heat flow (thermal energy AU per time interval A^), and where k 

W 

is the thermal conductivity (in units of ), A x is the area perpendicular to the heat 

mK 

flow, and where AT is the temperature difference across the material of width Ax . 



The temperature difference across the material can be measured either in degrees Kelvin 
or Celsius, since they are related one to the other by an additive value of 273 K. By 
practice, an engineer or technologists would use the familiar Celsius scale, while a 
scientists would tend to use the Kelvin scale. 



The values for the thermal conductivity of different materials is determined through 
experimental measurement and can be found in any good handbook or textbook. Some 
representative examples for solid materials is given in Table 1 , Thermal Conductivity of 
Solid Materials. 



Material 


W 
k ( ) 
mK 


Pure diamond 


3400 


Silver 


427 


Copper 


397 


Gold 


314 


Aluminium 


238 


Iron 


79.5 


Lead 


34.7 


Ice 


2.0 


Silica Brick 


1.1 


Standard Window Glass 


0.84 


Glass (pyrex) 


0.8 


Concrete 


0.8 


Compacted Snow 


0.2 


Wood 


0.08 


Loose Snow 


0.05 


Insulating Fibre Glass 


0.035 


Still Dry Air 


0.0234 


Table 1, Thermal Conductivity of Solid Materials 



You will note that a good thermal insulator is also a good electrical insulator. There is a 
theory that relates electron conductivity to thermal conductivity in good conductors that 
explains this, that you can search for and read about if you are at all interested. 



Example One: Conductive Heat Loss Through a Single 
Window Pane of Glass 

A simple example: of thermal conductivity is the conductive heat loss through a single 
window pane of glass with an area A = 1.0 m 2 , with a thickness of Ax = 0.005 m . For 

W 

standard window glass k lass = 0.84 . Let the interior building temperature Jj be 22 

mK 

C, and the exterior building temperature T 2 be C. 

A schematic of this system is given in Fig. 2. Conductive Heat Loss Through a Single 
Window Pane of Glass. 



T.=22C 




7\ = CC 



2a. = C.CC5m 



Fig. 2. Conductive Heat Loss Through a Single Window Pane of Glass 



The Temperature difference is then 



AT = T 2 -T l =(0-22)C 



By Fourier's law the conductive heat loss through this single window pane of glass is 

AT 



q x = -H 



Ax 



W ( 2 x(0-22) 

= -0.84 1.0m 2 p } - 

mK v } 0.005 m 

= 3,696 W 



which is a substantial amount of conductive heat that is leaving the inside of the building 
and is lost to the surrounding air (in comparison the radiative heat loss is around 74.6 
W, only 0.02 that of conductive heat loss). 

Example Two: The Igloo as a Practical Winter Shelter 

A second example of conductive heat loss is the simple Eskimo winter shelter made from 
compacted snow known as the igloo. In the classical 1922 film Nanook of the North you 
can watch an Eskimo family construct an igloo to protect against a night time temperature 
of-25C. 

The mean basal metabolism rate for a male adult Eskimo is around 10,000 kJ/day which 
translates into HOW of thermal heat while asleep. The Eskimo also use a stone lamp to 
light and heat the interior of the structure (producing around 300 W of thermal heat). 

In the film Nannook of the North the inhabitants of the igloo and their stone lamp is 
producing around 575 W of thermal energy, therefore 



q = -575 W 



Rewriting the Fourier Law and solving for the interior temperature 7J 



T =T + 



575W 


Ax 
A _ 


k 



Their igloo is a half sphere of about 2.25 metres in interior height, with compacted snow 
walls 0.25 m in thickness. Given its size and the fact it is a half-sphere the area of the 
outside surface of the igloo is A = 31.8 m 2 . The thermal conductivity of compacted 

W 



snow is A: , ,, = 0.20 

compacted snow 



mK 



Solving for the interior temperature we find that 



t; = - 25 c + 



575W 



0.20 

= - 25 C + 22.6 C 
= - 2.4 C 



0.25m 
31.8 m 2 



From a practical standpoint the interior temperature of the structure must remain below 
C to prevent the structure from melting and collapsing in on its occupants. If the exterior 
temperature drops to say - 50 C the Eskimo either make the walls of their igloo thicker 



and cover it with a measured layer of loose snow (& 



loose snow 



0.05 



W 
mK 



) or families 



double up and so there is twice as much body heat with the igloo itself. 



The igloo far from being a primitive structure is a simple albeit sophisticated example of 
building technology. For those of you who are winter out doors type, if you ever get lost 
or injured in the back country, you might build yourself a small snow home or igloo and 
wait out for the rescue (see for example the SAS Handbook). 



Composite Structures 

In our simple example of the single glaze window pane we haven't taken into account 
any heat loss through the window frame, which you may not could be significant if the 
frame is made out of a thermally conductive material like aluminium. 

To improve energy efficiency and to reduce heating costs it is common practice to use 
windows with two or more panes or glazes of glass to significantly reduce the heat loss. 
It is also common practice to make the window frames out of thermal insulating materials 
such as plastic or wood instead of aluminium. Note the value of k for aluminium 
compared to say wood (238:: 0.08) is a ratio of 3,000 to 1. 

These considerations leads to the question, how do we take into account such composite 
structures which are made with several materials with different thermal conductivity and 
thickness? It is also possible (see Question 3) to determine the temperature at an 
interface of materials by using the heat flux equation and solving for an indeterminate 
temperature T at the interface. 

Electric Analogy to Heat Transport by Conduction 

The treatment of heat transport by conduction is greatly simplified by using an analogy to 
current flow caused by a potential difference in a resistive material. This analogy been 
electrical and thermal flow has been confirmed experimentally and therefore can be used. 

In both the electrical and thermal systems 

Driving Force _ . 

= Resitance 

Flux 



Electrical Resistance: 

The driving force for current flow in an electrically conductive material is the potential 
difference AV . The Electrical Resistance of a material is a described by Ohm's Law, 



Voltage _AV 
Current I 



= R 



electrical 



For electrical resistors in series we recall that 



R Total - R \ +R 2+--^ R i 



and for resistors in parallel we recall that 



R 



1 11 ^1 

— = — + — + -" = 2j — 

Total ^1 ^2 i ^ 



Thermal Resistance: 

The driving force in thermal flow is the temperature difference AT . The Thermal 
Resistance of a material is described by 



AT 1 

q x ~k 



Ax 



= R 



Thermal 



What this means then is that 



AT 1 



q x k 



Ax 
~A 



= R 



AT 



Thermal 



R 



= q> 



Thermal 



We can use the basic theory of electrical resistors to also describe the thermal 
conductivity of composite structures. We can then determine the value for the rate of 
heat loss by first determining the total thermal resistance of the composite structure. 

Example Three: Double Pane of Glass 

Consider a composite structure made up of two panes of glass of area A = 1.0 m 2 each 
with the same thickness, with an air space of thickness Ax = 0.005 m between the panes. 
Let the interior building temperature Jj be 22 C, and the exterior building temperature T 2 
be C. A schematic is given in Fig. 2: Double Pane of Glass with Air Space. 



J = 21C 




r 2 = OC 



Fig. 2: Double Pane of Glass with Air Space 



The three layers are in series and so we can use the analogy of series resistance 



R Total -R l +R 2 +R 3 



1 


Ax l 


1 

+ — 

k 2 


Ax 2 
_A_ 


1 

+ — 

K 3 


Ax 3 

L4_ 




1 


"0.005m" 
_1.0m 2 _ 


1 

i 


o.s 


w 

?4 


0.0.234 


^ 



mK 
0.2256— 



0.005m 
1.0 m 2 



mK 



J 0.840.84 



m^T 



0.005m 
1.0 m 2 



Solving for the heat flow through the composite window we find 



AT 



R 



Total 

22K 



0.2256 



K 
W 



= 97.5 W 



Which is considerably less than the heat loss through a single pane of glass. 



Example Four: Heat Loss though an un-insulated Concrete 
Wall and a Single Pane Window. 

Consider a prefabricated wall 8.0 m x 5.0 m x 0.15 m thick made of un-insulated 
concrete. The wall has a 1.0 m x 1.0 m window made of single paned standard glass. Let 
the interior building temperature Jj be 22 C, and the exterior building temperature T 2 be 
OC. 

This composite system is an example of thermal resistance in parallel and as a results we 
must calculate the following, 



1 



R 



Total 



R 



+ 



1 



concrete 
1 



R 



f 



Ax 



N \ 



+ 



Window 
1 



( 



V 



hA) \hA 



Ax- 



N \ 



J 



11 _i_ 2^ 



Ax, Ax^ 



0.8 



V 



mK 



J 



7 ( W^ 

(39.0m 2 ) 0.84— (l.Ow 2 ) 



0.15m 

W W 

= 208 — + 168— 

K K 



+ ■ 



J 
0.005m 



= 376 



W_ 
K 



The heat flux out of the building is 



AT 



R 



Total 



= (22K)376 



= 8272 W 



W_ 
K 



A close look at this wall structure results in the following realization, that although the 
1.0 m 2 window represents only 1/40 1 of the entire wall area, it causes 45 % of the heat 
loss for the wall structure. 

Example Five: Heat Loss though an Insulated Concrete Wall 
and a Single Pane Window. 



If we decide to leave the window as is, how much fibre glass insulated would have to be 
installed to reduce the heat loss by 45% to compensate for the heat loss through the 
window? Let us decide that we want to add 0.10 m of fibre glass with 

W 



k mre Glass = 0.035 



mK 



R 



Insulated Wall 



Concrete +&**_<*. 

0.15m 



f 



W 



\ 



+ 



0.10m 



0.8 
V mK j 



(39.0m 2 ) 



f 



0.035 



W 



\ 



V 



mK j 



(39.0m 2 ) 



= 0.00481 — + 0.0733 — 
W W 

= 0.0781 — 
W 



For the insulated concrete wall with insulation, 



1 ' =12.81 W 



r K ' i<r 

iV Insulated_Wall 0733 

w 

The total thermal resistance for the insulated wall and single glazed window is 

1 1 1 

= + 



D D D 

Total Insulated _ Wall Window 

w w 

= 12.81 — + 168 — 
K K 

W 

= 180.8 — 

K 

The Heat Flux for the insulated wall and single glazed window is 



AT 
q * = R 

^ Total 

= (22K)180.8 — 
= 3978 W 

which is about half of the heat loss through an un-insulated concrete wall with single 
pane window. 



Example Six: Heat Loss though an Un-insulated Concrete 
Wall and a Double Pane Window. 

Consider a prefabricated wall 8.0 m x 5.0 m x 0.15 m thick made of un-insulated 
concrete. The wall has a 1.0 m x 1.0 m window made of double paned standard glass (to 
simplify our example we shall use the result from example three for the window). Let the 
interior building temperature Jj be 22 C, and the exterior building temperature T 2 be C. 

This composite system is an example of thermal resistance in parallel and as a results we 
must calculate the following, 



1 



1 



R 



Total 



R 



■ + • 



1 



concrete 
1 



R 



+ 



Window 
1 



r Ax^ 



V 



Mi 

M 

Ax l 
f 



0.2256 



J 



K_ 
W 



+ 4.433 



W_ 
K 



W 
0.8 — 
V mK 



\ 



(39.0m 2 ) 



J 



0.15m 

W W 

= 208 — + 4.433 — 

K K 



+ 4.433 



W_ 
K 



= 212.433 



W_ 

K 



The heat flux for the wall with a double glazed window is 



AT 



-^ Total 



= (22K)212.433 — 
= 4673.5 W 

As you see, it pays to put in better windows for it means less heat loss through the wall 
structure, better energy efficiency and lower heating costs during the colder months of the 
year. It is easy to convince yourself that during the warmer months of the year the better 
wall system also reduces the cost to cool such a structure as well. 

Example Seven: Heat Loss though an Insulated Concrete Wall 
and a Double Pane Window. 

It is straightforward to calculate the heat loss in the case of an insulated concrete wall 
with a double glazed window pane, it is merely 



AT 



^ Total 

= (22K)17.243 — 
= 379.3 W 

Which is 1/20 of that of the un-insulated concrete with a single pane window. 



Homework Questions: 

1 . You are by yourself and lost during a cross country skiing trip in the back country of 
British Columbia. The sun is going down and the temperature is about to drop to - 40 C. 
You remember learning about igloos in your BCIT physics course and you decide to 
build yourself a snow structure to wait out the night and rescue. Before you retire for 
the night you build a fire out side the structure far enough away that you do not melt the 
igloo and bright enough to be seen a good distant away. 

Design a snow structure out of loose and compact snow that will keep the inner 
temperature of your refuge at no colder than -2.0 C. Proof your design by calculation. 

2. As mentioned, it is common practice to make the window frames out of thermal 
insulating materials such as plastic or wood, instead of aluminium. Note the value of k 
for aluminium compared to say wood (238 :: 0.08) is a ratio of 3,000 to 1. Estimate the 
heat loss caused by an aluminium window frame for a 1 .0 m single pane glass window. 

3. Using what you know about heat flux through a composite structure derive an 
expression for temperature in a wall prove that the temperature at an interface between 
two materials in a wall is 

_ k { Ax 2 T { + k 2 Ax { T 2 

k { Ax 2 + k 2 Ax 1 

The vapour barrier in a wall is a plastic film designed to keep moisture from condensing 

W 

on the cold outer structure of a wall. If T x = 22C and T 2 = -25C , k x = 0.035 , (fibre 

mK 

W 

glass insulation) and & 2 =0.08 (wood exterior) and Ax 2 = 0.15m, estimate the 

mK 

minimum distance for Ax l to keep the interface or vapour barrier above C.